3.8.19 \(\int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=30 \[ \frac {2 \sqrt {a+b x}}{\sqrt {c+d x} (b c-a d)} \]

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Rubi [A]  time = 0.00, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {37} \begin {gather*} \frac {2 \sqrt {a+b x}}{\sqrt {c+d x} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b*x]*(c + d*x)^(3/2)),x]

[Out]

(2*Sqrt[a + b*x])/((b*c - a*d)*Sqrt[c + d*x])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx &=\frac {2 \sqrt {a+b x}}{(b c-a d) \sqrt {c+d x}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 30, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {a+b x}}{\sqrt {c+d x} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + b*x]*(c + d*x)^(3/2)),x]

[Out]

(2*Sqrt[a + b*x])/((b*c - a*d)*Sqrt[c + d*x])

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IntegrateAlgebraic [A]  time = 0.00, size = 30, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {a+b x}}{\sqrt {c+d x} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(Sqrt[a + b*x]*(c + d*x)^(3/2)),x]

[Out]

(2*Sqrt[a + b*x])/((b*c - a*d)*Sqrt[c + d*x])

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fricas [A]  time = 1.38, size = 42, normalized size = 1.40 \begin {gather*} \frac {2 \, \sqrt {b x + a} \sqrt {d x + c}}{b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(1/2)/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

2*sqrt(b*x + a)*sqrt(d*x + c)/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x)

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giac [A]  time = 1.23, size = 47, normalized size = 1.57 \begin {gather*} \frac {2 \, \sqrt {b x + a} b^{2}}{\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (b c {\left | b \right |} - a d {\left | b \right |}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(1/2)/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

2*sqrt(b*x + a)*b^2/(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(b*c*abs(b) - a*d*abs(b)))

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maple [A]  time = 0.00, size = 27, normalized size = 0.90 \begin {gather*} -\frac {2 \sqrt {b x +a}}{\sqrt {d x +c}\, \left (a d -b c \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(1/2)/(d*x+c)^(3/2),x)

[Out]

-2*(b*x+a)^(1/2)/(d*x+c)^(1/2)/(a*d-b*c)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(1/2)/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [B]  time = 1.31, size = 26, normalized size = 0.87 \begin {gather*} -\frac {2\,\sqrt {a+b\,x}}{\left (a\,d-b\,c\right )\,\sqrt {c+d\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(1/2)*(c + d*x)^(3/2)),x)

[Out]

-(2*(a + b*x)^(1/2))/((a*d - b*c)*(c + d*x)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a + b x} \left (c + d x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(1/2)/(d*x+c)**(3/2),x)

[Out]

Integral(1/(sqrt(a + b*x)*(c + d*x)**(3/2)), x)

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